SOLUTION: Suppose that the sitting​ back-to-knee length for a group of adults has a normal distribution with a mean of mu=22.9in. and a standard deviation of alpha= 1.2in. These data are o

Algebra ->  Probability-and-statistics -> SOLUTION: Suppose that the sitting​ back-to-knee length for a group of adults has a normal distribution with a mean of mu=22.9in. and a standard deviation of alpha= 1.2in. These data are o      Log On


   



Question 1177857: Suppose that the sitting​ back-to-knee length for a group of adults has a normal distribution with a mean of mu=22.9in. and a standard deviation of alpha= 1.2in. These data are often used in the design of different​ seats, including aircraft​ seats, train​ seats, theater​ seats, and classroom seats. Instead of using 0.05 for identifying significant​ values, use the criteria that a value x is significantly high if​ P(x or ​greater)0.01 and a value is significantly low if​ P(x or ​less)0.01. Find the​ back-to-knee lengths separating significant values from those that are not significant. Using these​ criteria, is a​ back-to-knee length of 25.2in. significantly​ high?
Answer by CPhill(1987) About Me  (Show Source):
You can put this solution on YOUR website!
Let's solve this problem step-by-step.
**1. Find the Z-scores for the Significant Values**
* We are given that a value is significantly low if P(x or less) < 0.01.
* We are given that a value is significantly high if P(x or greater) < 0.01.
To find the back-to-knee lengths that separate significant values, we need to find the z-scores corresponding to the 0.01 and 0.99 percentiles of the standard normal distribution.
* For the lower bound (0.01 percentile), we need to find the z-score such that P(Z < z) = 0.01.
* Using a standard normal distribution table or a calculator, we find that z_lower ≈ -2.33.
* For the upper bound (0.99 percentile), we need to find the z-score such that P(Z > z) = 0.01, or equivalently, P(Z < z) = 0.99.
* Using a standard normal distribution table or a calculator, we find that z_upper ≈ 2.33.
**2. Calculate the Back-to-Knee Lengths**
We are given:
* Mean (μ) = 22.9 inches
* Standard deviation (σ) = 1.2 inches
We use the formula:
* x = μ + z * σ
* Lower bound:
* x_lower = 22.9 + (-2.33) * 1.2
* x_lower = 22.9 - 2.796
* x_lower ≈ 20.104 inches
* Upper bound:
* x_upper = 22.9 + (2.33) * 1.2
* x_upper = 22.9 + 2.796
* x_upper ≈ 25.696 inches
So, the back-to-knee lengths separating significant values are approximately 20.10 inches (lower bound) and 25.70 inches (upper bound).
**3. Determine if 25.2 Inches is Significantly High**
* We are asked if a back-to-knee length of 25.2 inches is significantly high.
* We found that the upper bound for significantly high values is approximately 25.70 inches.
* Since 25.2 inches < 25.70 inches, a back-to-knee length of 25.2 inches is not significantly high.
**Results**
* Lower bound for significant values: ≈ 20.10 inches
* Upper bound for significant values: ≈ 25.70 inches
* A back-to-knee length of 25.2 inches is not significantly high.